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# Solucionario serway cap 35

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### Solucionario serway cap 35

1. 1. 35 The Nature of Light and the Laws of Geometric Optics CHAPTER OUTLINE 35.1 The Nature of Light 35.2 Measurements of the Speed of Light 35.3 The Ray Approximation in Geometric Optics 35.4 The Wave Under Reﬂection 35.5 The Wave Under Refraction 35.6 Huygens’s Principle 35.7 Dispersion 35.8 Total Internal Reﬂection ANSWERS TO QUESTIONS Q35.1 Light travels through a vacuum at a speed of 300 000 km per second. Thus, an image we see from a distant star or galaxy must have been generated some time ago. For example, the star Altair is 16 light-years away; if we look at an image of Altair today, we know only what was happening 16 years ago. This may not initially seem signiﬁcant, but astronomers who look at other galaxies can gain an idea of what galaxies looked like when they were signiﬁcantly younger. Thus, it actually makes sense to speak of “looking backward in time.” *Q35.2 104 m/(3 × 108 m/s) is 33 µs. Answer (c). *Q35.3 We consider the quantity l /d. The smaller it is, the better the ray approximation works. In (a) it is like 0.34 m/1 m ≈ 0.3. In (b) we can have 0.7 µm/2 mm ≈ 0.000 3. In (c), 0.4 µm/2 mm ≈ 0.000 2. In (d), 300 m/1 m ≈ 300. In (e) 1 nm/1 mm ≈ 0.000 001. The ranking is then e, c, b, a, d. Q35.4 With a vertical shop window, streetlights and his own reﬂection can impede the window shopper’s clear view of the display. The tilted shop window can put these reﬂections out of the way. Windows of airport control towers are also tilted like this, as are automobile windshields. 295 FIG. Q35.4
2. 2. 296 Chapter 35 Q35.5 We assume that you and the child are always standing close together. For a ﬂat wall to make an echo of a sound that you make, you must be stand- ing along a normal to the wall. You must be on the order of 100 m away, to make the transit time sufﬁciently long that you can hear the echo sepa- rately from the original sound. Your sound must be loud enough so that you can hear it even at this consider- able range. In the picture, the dashed rectangle represents an area in which you can be standing. The arrows represent rays of sound. Now suppose two vertical perpendicular walls form an inside corner that you can see. Some of the sound you radiate horizontally will be headed generally toward the corner. It will reﬂect from both walls with high efﬁciency to reverse in direction and come back to you. You can stand anywhere reasonably far away to hear a retroreﬂected echo of sound you produce. If the two walls are not perpen- dicular, the inside corner will not produce retroreﬂection. You will generally hear no echo of your shout or clap. If two perpendicular walls have a reasonably narrow gap between them at the corner, you can still hear a clear echo. It is not the corner line itself that retroreﬂects the sound, but the perpendicular walls on both sides of the corner. Diagram (b) applies also in this case. Q35.6 The stealth ﬁghter is designed so that adjacent panels are not joined at right angles, to prevent any retroreﬂection of radar signals. This means that radar signals directed at the ﬁghter will not be channeled back toward the detector by reﬂection. Just as with sound, radar signals can be treated as diverging rays, so that any ray that is by chance reﬂected back to the detector will be too weak in intensity to distinguish from background noise. This author is still waiting for the automotive industry to utilize this technology. *Q35.7 Snell originally stated his law in terms of cosecants. From v = c/n and sinθ = 1/cscθ and λ = c/nf with c and f constant between media, we conclude that a, b, and c are all correct statements. Q35.8 An echo is an example of the reﬂection of sound. Hearing the noise of a distant highway on a cold morning, when you cannot hear it after the ground warms up, is an example of acoustical refraction. You can use a rubber inner tube inﬂated with helium as an acoustical lens to concen- trate sound in the way a lens can focus light. At your next party, see if you can experimentally ﬁnd the approximate focal point! 296 Chapter 35 (a) (b) FIG. Q35.5
4. 4. 298 Chapter 35 *Q35.18 Answer (c). We want a big difference between indices of refraction to have total internal reﬂec- tion under the widest range of conditions. Q35.19 A mirage occurs when light changes direction as it moves between batches of air having different indices of refraction because they have different densities at different temperatures. When the sun makes a blacktop road hot, an apparent wet spot is bright due to refraction of light from the bright sky. The light, originally headed a little below the horizontal, always bends up as it ﬁrst enters and then leaves sequentially hotter, lower-density, lower-index layers of air closer to the road surface. Section 35.1 The Nature of Light Section 35.2 Measurements of the Speed of Light *P35.1 The Moon’s radius is 1 74 106 . × m and the Earth’s radius is 6 37 106 . × m. The total distance traveled by the light is: d = × − × − ×( )= ×2 3 84 10 1 74 10 6 37 10 7 528 6 6 . . . .m m m 1108 m This takes 2.51 s, so v = × = × = 7 52 10 2 995 10 299 5 8 8. . . m 2.51 s m s Mm s . The sizes of the objects need to be taken into account. Otherwise the answer would be too large by 2%. P35.2 ∆x ct= ; c x t = = ×( )( ) ( ) ∆ 2 1 50 10 1 000 22 0 60 8 . . . km m km min 00 2 27 10 2278 s min m s Mm s ( ) = × =. P35.3 The experiment is most convincing if the wheel turns fast enough to pass outgoing light through one notch and returning light through the next: t c = 2 θ ω ω= = ⎛ ⎝⎜ ⎞ ⎠⎟t c 2 so ω θ π = = ×( ) ( )⎡⎣ ⎤⎦ ×( ) = c 2 2 998 10 2 720 2 11 45 10 1 8 3 . . 114 rad s The returning light would be blocked by a tooth at one-half the angular speed, giving another data point. SOLUTIONS TO PROBLEMS
5. 5. The Nature of Light and the Laws of Geometric Optics 299 Section 35.3 The Ray Approximation in Geometric Optics Section 35.4 The Wave Under Reﬂection Section 35.5 The Wave Under Refraction P35.4 (a) Let AB be the originally horizontal ceiling, BC its originally vertical normal, AD the new ceiling, and DE its normal. Then angle BAD = φ. By deﬁnition DE is perpendicular to AD and BC is perpendicular to AB. Then the angle between DE extended and BC is φ because angles are equal when their sides are perpendicular, right side to right side and left side to left side. (b) Now CBE = φ is the angle of incidence of the vertical light beam. Its angle of reﬂection is also φ. The angle between the vertical inci- dent beam and the reﬂected beam is 2φ. (c) tan . .2 1 40 0 001 94φ = = cm 720 cm φ = °0 055 7. P35.5 (a) From geometry, 1 25 40 0. sin .m = °d so d = 1 94. m (b) 50 0. ° above the horizontal or parallel to the incident ray. P35.6 (a) Method One: The incident ray makes angle α θ= ° −90 1 with the ﬁrst mirror. In the picture, the law of reﬂection implies that θ θ1 1= ′ Then β θ θ α= ° − ′ = − =90 901 1 In the triangle made by the mirrors and the ray passing between them, β γ γ β + ° + = ° = ° − 90 180 90 Further, δ γ β α= ° − = =90 and ∈= =δ α Thus the ﬁnal ray makes the same angle with the ﬁrst mirror as did the incident ray. Its direction is opposite to the incident ray. Method Two: The vector velocity of the incident light has a component vy perpendicular to the ﬁrst mirror and a component vx perpendicular to the second. The vy component is reversed upon the ﬁrst Mirror 2 50° 40° 50° 50° 40° d i1 = 40° i2 = 50° Mirror 11.25 m 40.0° 1.25 m Mirror 1 Mirror 2 Light beam P FIG. P35.5 A D E FIG. P35.4 (b) A B C D E FIG. P35.4 (a) FIG. P35.6 continued on next page
6. 6. 300 Chapter 35 reﬂection, which leaves vx unchanged. The second reﬂection reverses vx and leaves vy unchanged. The doubly reﬂected ray then has velocity opposite to the incident ray. (b) The incident ray has velocity v v vx y z ˆ ˆ ˆi j k+ + . Each reﬂection reverses one component and leaves the other two unchanged. After all the reﬂections, the light has velocity − − −v v vx y z ˆ ˆ ˆi j k, opposite to the incident ray. P35.7 Let d represent the perpendicular distance from the person to the mirror. The distance between lamp and person measured parallel to the mirror can be written in two ways: 2d d dtan tan tanθ θ φ+ = . The condition on the distance traveled by the light is 2 2d d d cos cos cosφ θ θ = + . We have the two equations 3tan tanθ φ= and 2 3cos cosθ φ= . To eliminate φ we write 9 2 2 2 2 sin cos sin cos θ θ φ φ = 4 92 2 cos cosθ φ= 9 1 4 2 2 2 2 2 2 2 cos sin cos cos cos sin cos φ θ θ φ θ θ θ = −( ) = 11 4 9 4 1 4 9 1 36 9 2 2 2 2 −⎛ ⎝ ⎞ ⎠ = − −( ) = cos sin sin sin θ θ θ θ −− + = = ° 4 4 5 32 23 3 2 2 sin sin . θ θ θ *P35.8 The excess time the second pulse spends in the ice is 6.20 m/[(3.00 × 108 m/s)/1.309] = 27.1 ns P35.9 Using Snell’s law, sin sinθ θ2 1 2 1= n n θ2 25 5= °. λ λ 2 1 1 442= = n nm 2d d d 2d tan q d tan q d tan f f q q FIG. P35.7 FIG. P35.9
7. 7. The Nature of Light and the Laws of Geometric Optics 301 *P35.10 The law of refraction n n1 1 2 2sin sinθ θ= can be put into the more general form c c v v v v 1 1 2 2 1 1 2 2 sin sin sin sin θ θ θ θ = = In this form it applies to all kinds of waves that move through space. sin . sin sin . . 3 5 343 1 493 0 266 15 2 2 2 ° = = = m s m s θ θ θ 44° The wave keeps constant frequency in f = = = = ( ) v v v v 1 1 2 2 2 2 1 1 1 493 0 589 343 λ λ λ λ m s m m s . == 2 56. m The light wave slows down as it moves from air into water but the sound speeds up by a large factor. The light wave bends toward the normal and its wavelength shortens, but the sound wave bends away from the normal and its wavelength increases. P35.11 n n1 1 2 2sin sinθ θ= sin . sin sin . . . θ θ θ 1 1 1 333 45 1 33 0 707 0 943 = ° = ( )( ) = 11 70 5 19 5= ° → °. . above the horizon P35.12 (a) f c = = × × = ×− λ 3 00 10 6 328 10 4 74 10 8 7 14. . . m s m Hz (b) λ λ glass air nm 1.50 nm= = = n 632 8 422 . (c) vglass air m s m s= = × = × = c n 3 00 10 1 50 2 00 10 200 8 8. . . MMm s P35.13 We ﬁnd the angle of incidence: n n1 1 2 2 1 1 1 333 1 52 19 6 22 sin sin . sin . sin . θ θ θ θ = = ° = ..5° The angle of reﬂection of the beam in water is then also 22 5. ° . FIG. P35.11
8. 8. 302 Chapter 35 *P35.14 (a) As measured from the diagram, the incidence angle is 60°, and the refraction angle is 35°. From Snell’s law, sin sin θ θ 2 1 2 1 = v v , then sin sin 35 60 2° ° = v c and the speed of light in the block is 2 0 108 . × m s . (b) The frequency of the light does not change upon refraction. Knowing the wavelength in a vacuum, we can use the speed of light in a vacuum to determine the frequency: c f= λ, thus 3 00 10 632 8 108 9 . .× = ×( )− f , so the frequency is 474 1. THz . (c) To ﬁnd the wavelength of light in the block, we use the same wave speed relation, v = f λ , so 2 0 10 4 741 108 14 . .× = ×( )λ , so λglass nm= 420 . P35.15 (a) Flint Glass: v = = × = × = c n 3 00 10 1 66 1 81 10 181 8 8. . . m s m s Mm s (b) Water: v = = × = × = c n 3 00 10 1 333 2 25 10 225 8 8. . . m s m s Mm s (c) Cubic Zirconia: v = = × = × = c n 3 00 10 2 20 1 36 10 136 8 8. . . m s m s Mm s *P35.16 From Snell’s law, sin sin .θ = ⎛ ⎝⎜ ⎞ ⎠⎟ ° n n medium liver 50 0 But n n c c medium liver medium liver liver medium = = v v v v == 0 900. so θ = ( ) °⎡⎣ ⎤⎦ = °− sin . sin . .1 0 900 50 0 43 6 From the law of reﬂection, d = = 12 0 6 00 . . cm 2 cm, and h d = = °( ) = tan . tan . . θ 6 00 43 6 6 30 cm cm P35.17 n n1 1 2 2sin sinθ θ= : θ θ 2 1 1 1 2 = ⎛ ⎝⎜ ⎞ ⎠⎟ − sin sinn n θ2 1 1 00 30 1 50 19 5= °⎧ ⎨ ⎩ ⎫ ⎬ ⎭ = °− sin . sin . . θ2 and θ3 are alternate interior angles formed by the ray cutting parallel normals. So, θ θ3 2 19 5= = °. 1 50 1 00 30 0 3 4 4 . sin . sin . θ θ θ = = ° FIG. P35.17 50° 12 cm Tumor d h nmedium nliver q FIG. P35.16
9. 9. The Nature of Light and the Laws of Geometric Optics 303 P35.18 sin sinθ θ1 2= nw sin . sin . sin . . .θ θ2 1 1 1 333 1 1 333 90 0 28 0 0= = ° − °( ) = 6662 0 662 41 5 3 00 2 1 2 θ θ = ( ) = ° = = − sin . . tan . h d m tan441.5 m ° = 3 39. P35.19 At entry, n n1 1 2 2sin sinθ θ= or 1 00 30 0 1 50 2. sin . . sin° = θ θ2 19 5= °. The distance h the light travels in the medium is given by cos . θ2 2 00 = cm h or h = ° = 2 00 2 12 . . cm cos19.5 cm The angle of deviation upon entry is α θ θ= − = ° − ° = °1 2 30 0 19 5 10 5. . . The offset distance comes from sinα = d h : d = ( ) ° =2 21 10 5 0 388. sin . .cm cm P35.20 The distance h traveled by the light is h = ° = 2 00 2 12 . . cm cos19.5 cm The speed of light in the material is v = = × = × c n 3 00 10 1 50 2 00 10 8 8. . . m s m s Therefore, t h = = × × = × = − − v 2 12 10 1 06 10 10 2 10. . m 2.00 10 m s s8 66 ps P35.21 Applying Snell’s law at the air-oil interface, n nair oilsin sin .θ = °20 0 yields θ = °30 4. Applying Snell’s law at the oil-water interface n nw sin sin .′ = °θ oil 20 0 yields ′ = °θ 22 3. water n = 1.333 air n = 1.00 28° 3.0 m h q1 = 62° q2 FIG. P35.18 FIG. P35.21 FIG. P35.19
10. 10. 304 Chapter 35 P35.22 For sheets 1 and 2 as described, n n n n 1 2 1 2 26 5 31 7 0 849 sin . sin . . ° = ° = For the trial with sheets 3 and 2, n n n n 3 2 3 2 26 5 36 7 0 747 sin . sin . . ° = ° = Now 0 747 0 849 1 14 3 1 3 1 . . . n n n n = = For the third trial, n n n1 3 3 1 3 3 26 5 1 14 23 1 sin . sin . sin . ° = = = ° θ θ θ *P35.23 Refraction proceeds according to 1 00 1 661 2. sin . sin( ) = ( )θ θ (1) (a) For the normal component of velocity to v v1 1 2 2cos cosθ θ= be constant, or c c ( ) = ⎛ ⎝⎜ ⎞ ⎠⎟cos . cosθ θ1 2 1 66 (2) We multiply Equations (1) and (2), obtaining: sin cos sin cosθ θ θ θ1 1 2 2= or sin sin2 21 2θ θ= The solution θ θ1 2 0= = does not satisfy Equation (2) and must be rejected. The physical solution is 2 180 21 2θ θ= ° − or θ θ2 190 0= ° −. . Then Equation (1) becomes: sin . cosθ θ1 11 66= , or tan .θ1 1 66= which yields θ1 58 9= °. In this case, yes , the perpendicular velocity component does remain constant. (b) Light entering the glass slows down and makes a smaller angle with the normal. Both effects reduce the velocity component parallel to the surface of the glass. Then no, the parallel velocity component cannot rremain constant, or will remain constant onnly in the trivial case θ θ1 2 0= = .
11. 11. The Nature of Light and the Laws of Geometric Optics 305 P35.24 Consider glass with an index of refraction of 1.5, which is 3 mm thick. The speed of light in the glass is 3 10 1 5 2 10 8 8× = × m s m s . The extra travel time is 3 10 2 10 3 10 3 10 10 3 8 3 8 11× × − × × − − −m m s m m s s~ For light of wavelength 600 nm in vacuum and wavelength 600 400 nm 1.5 nm= in glass, the extra optical path, in wavelengths, is 3 10 3 10 6 10 10 3 7 3 7 3× × − × × − − − − m 4 10 m m m wavel~ eengths P35.25 Taking Φ to be the apex angle and δmin to be the angle of minimum deviation, from Equation 35.9, the index of refraction of the prism material is n = +( )⎡⎣ ⎤⎦ ( ) sin sin minΦ Φ δ 2 2 Solving for δmin, δmin sin sin sin . sin= ⎛ ⎝⎜ ⎞ ⎠⎟ − = ( )− − 2 2 2 2 20 251 1 n Φ Φ .. . .0 50 0 86 8°( )⎡⎣ ⎤⎦ − ° = ° P35.26 n 700 1 458nm( ) = . (a) 1 00 75 0 1 458 2. sin . . sin( ) ° = θ ; θ2 41 5= °. (b) Let θ β3 90 0+ = °. , θ α2 90 0+ = °. then α β+ + ° = °60 0 180. So 60 0 0 60 0 41 5 18 52 3 3. . . .° − − = ⇒ ° − ° = = °θ θ θ (c) 1 458 18 5 1 00 4. sin . . sin° = θ θ4 27 6= °. (d) γ θ θ β θ= −( )+ − ° −( )⎡⎣ ⎤⎦1 2 490 0. γ = ° − ° + ° − °( )− ° − °( ) =75 0 41 5 90 0 18 5 90 0 27 6 4. . . . . . 22 6. ° P35.27 At the ﬁrst refraction, 1 00 1 2. sin sinθ θ= n The critical angle at the second surface is given by nsin .θ3 1 00= : or θ3 1 1 00 1 50 41 8= ⎛ ⎝⎜ ⎞ ⎠⎟ = °− sin . . . But, θ θ2 360 0= ° −. Thus, to avoid total internal reﬂection at the second surface (i.e., have θ3 41 8< °. ) it is necessary that θ2 18 2> °. Since sin sinθ θ1 2= n , this becomes sin . sin . .θ1 1 50 18 2 0 468> ° = or θ1 27 9> °. 60.0° γα q1 q2 q3 q4 β FIG. P35.26 FIG. P35.27
12. 12. 306 Chapter 35 P35.28 At the ﬁrst refraction, 1 00 1 2. sin sinθ θ= n The critical angle at the second surface is given by nsin .θ3 1 00= , or θ3 1 1 00 = ⎛ ⎝⎜ ⎞ ⎠⎟ − sin . n But 90 0 90 0 1802 3. .° −( )+ ° −( )+ = °θ θ Φ which gives θ θ2 3= −Φ Thus, to have θ3 1 1 00 < ⎛ ⎝⎜ ⎞ ⎠⎟ − sin . n and avoid total internal reﬂection at the second surface, it is necessary that θ2 1 1 00 > − ⎛ ⎝⎜ ⎞ ⎠⎟ − Φ sin . n Since sin sinθ θ1 2= n , this requirement becomes sin sin sin . θ1 1 1 00 > − ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − n n Φ or θ1 1 1 1 00 > − ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⎛ ⎝⎜ ⎞ ⎠⎟ − − sin sin sin . n n Φ Through the application of trigonometric identities, θ1 1 2 1> − −( )− sin sin cosn Φ Φ P35.29 Note for use in every part: Φ + ° −( )+ ° −( ) = °90 0 90 0 1802 3. .θ θ so θ θ3 2= −Φ At the ﬁrst surface the deviation is α θ θ= −1 2 At exit, the deviation is β θ θ= −4 3 The total deviation is therefore δ α β θ θ θ θ θ θ= + = + − − = + −1 4 2 3 1 4 Φ (a) At entry: n n1 1 2 2sin sinθ θ= or θ2 1 48 6 1 50 30 0= °⎛ ⎝⎜ ⎞ ⎠⎟ = °− sin sin . . . Thus, θ3 60 0 30 0 30 0= ° − ° = °. . . At exit: 1 50 30 0 1 00 4. sin . . sin° = θ or θ4 1 1 50 30 0 48 6= °( )⎡⎣ ⎤⎦ = °− sin . sin . . so the path through the prism is symmetric whenθ1 48 6= °. . (b) δ = ° + ° − ° = °48 6 48 6 60 0 37 2. . . . (c) At entry: sin sin . . .θ θ2 2 45 6 1 50 28 4= ° ⇒ = ° θ3 60 0 28 4 31 6= ° − ° = °. . . At exit: sin . sin . .θ θ4 41 50 31 6 51 7= °( )⇒ = ° δ = ° + ° − ° = °45 6 51 7 60 0 37 3. . . . Φ q2 q1 q3 FIG. P35.28 FIG. P35.29 continued on next page
13. 13. The Nature of Light and the Laws of Geometric Optics 307 (d) At entry: sin sin . . .θ θ2 2 51 6 1 50 31 5= ° ⇒ = ° θ3 60 0 31 5 28 5= ° − ° = °. . . At exit: sin . sin . .θ θ4 41 50 28 5 45 7= °( )⇒ = ° δ = ° + ° − ° = °51 6 45 7 60 0 37 3. . . . Section 35.6 Huygens’s Principle P35.30 (a) For the diagrams of contour lines and wave fronts and rays, see Figures (a) and (b) below. As the waves move to shallower water, the wave fronts bend to become more nearly parallel to the contour lines. (b) For the diagrams of contour lines and wave fronts and rays, see Figures (c) and (d) below. We suppose that the headlands are steep underwater, as they are above water. The rays are everywhere perpendicular to the wave fronts of the incoming refracting waves. As shown, the rays bend toward the headlands and deliver more energy per length at the headlands. FIG. P35.30 Section 35.7 Dispersion P35.31 For the incoming ray, sin sin θ θ 2 1 = n Using the ﬁgure to the right, θ2 1 50 0 1 66 27 48( ) = °⎛ ⎝⎜ ⎞ ⎠⎟ = °− violet sin sin . . . θ2 1 50 0 1 62 28 22( ) = °⎛ ⎝⎜ ⎞ ⎠⎟ = °− red sin sin . . . For the outgoing ray, θ θ3 260 0= ° −. and sin sinθ θ4 3= n : θ4 1 1 66 32 52 63 17( ) = °[ ]= °− violet sin . sin . . θ4 1 1 62 31 78 58 56( ) = °[ ]= °− red sin . sin . . The angular dispersion is the difference ∆θ θ θ4 4 4 63 17 58 56 4 61= ( ) − ( ) = ° − ° = °violet red . . . FIG. P35.31
14. 14. 308 Chapter 35 P35.32 From Fig 35.21 nv = 1 470. at 400 nm and nr = 1 458. at 700 nm Then 1 00 1 470. sin . sinθ θ= v and 1 00 1 458. sin . sinθ θ= r δ δ θ θ θ θ r r− = − = ⎛ ⎝ ⎞ ⎠ −− − v v sin sin . sin sin . 1 1 1 458 1 4770 30 0 1 458 301 1 ⎛ ⎝ ⎞ ⎠ = °⎛ ⎝ ⎞ ⎠ −− − ∆δ sin sin . . sin sin .. . . 0 1 470 0 171 °⎛ ⎝ ⎞ ⎠ = ° Section 35.8 Total Internal Reﬂection P35.33 nsinθ = 1. From Table 35.1, (a) θ = ⎛ ⎝⎜ ⎞ ⎠⎟ = °− sin . .1 1 2 419 24 4 (b) θ = ⎛ ⎝⎜ ⎞ ⎠⎟ = °− sin . .1 1 1 66 37 0 (c) θ = ⎛ ⎝⎜ ⎞ ⎠⎟ = °− sin . .1 1 1 309 49 8 *P35.34 For total internal reﬂection, n n1 1 2 90 0sin sin .θ = ° 1 50 1 33 1 001. sin . ( . )θ = or θ1 62 5= °. P35.35 sinθc n n = 2 1 n n2 1 88 8 1 000 3 0 999 8 1 000 08= ° = ( )( ) =sin . . . . P35.36 sin . . .θc n n = = =air pipe 1 00 1 36 0 735 θc = °47 3. Geometry shows that the angle of refraction at the end is φ θ= ° − = ° − ° = °90 0 90 0 47 3 42 7. . . .c Then, Snell’s law at the end, 1 00 1 36 42 7. sin . sin .θ = ° gives θ = °67 2. The 2-mm diameter is unnecessary information. FIG. P35.35 FIG. P35.36
15. 15. The Nature of Light and the Laws of Geometric Optics 309 P35.37 (a) A ray along the inner edge will escape if any ray escapes. Its angle of incidence is described by sinθ = −R d R and by nsin sinθ > °1 90 . Then n R d R −( ) > 1 nR nd R− > nR R nd− > R nd n > −1 (b) As d → 0, Rmin → 0. This is reasonable. As n increases, Rmin decreases. This is reasonable. As n decreases toward 1, Rmin increases. This is reasonable. (c) Rmin . . = ×( ) = × − − 1 40 100 10 0 40 350 10 6 6 m m P35.38 (a) sin sin θ θ 2 1 2 1 = v v and θ2 90 0= °. at the critical angle sin . sin 90 0 1 850 343 ° = θc m s m s so θc = ( ) = °− sin . .1 0 185 10 7 (b) Sound can be totally reﬂected if it is traveling in the medium where it travels slower: air . (c) Sound in air falling on the wall from most ddirections is 100% reflected , so the wall is a good mirror. P35.39 For plastic with index of refraction n ≥ 1 42. surrounded by air, the critical angle for total inter- nal reﬂection is given by θc n = ⎛ ⎝ ⎞ ⎠ ≤ ⎛ ⎝ ⎞ ⎠ = °− − sin sin . .1 11 1 1 42 44 8 In the gasoline gauge, skylight from above travels down the plastic. The rays close to the vertical are totally reﬂected from the sides of the slab and from both facets at the lower end of the plastic, where it is not immersed in gasoline. This light returns up inside the plastic and makes it look bright. Where the plastic is immersed in gasoline, with index of refraction about 1.50, total internal reﬂection should not happen. The light passes out of the lower end of the plastic with little reﬂected, making this part of the gauge look dark. To frustrate total internal reﬂection in the gasoline, the index of refraction of the plastic should be n < 2 12. . since θc = ⎛ ⎝⎜ ⎞ ⎠⎟ = °− sin . . .1 1 50 2 12 45 0 Additional Problems *P35.40 From the textbook ﬁgure we have w b a= +2 so b w a b t = − = − = = = 2 700 1 349 5 349 5 2 m m 2 m µ µ µ θ . tan . µµ µ θ m 1 200 m = = °0 291 16 22. . For refraction at entry, n n n n 1 1 2 2 1 1 2 2 1 1 1 55 sin sin sin sin sin . s θ θ θ θ = = =− − iin . . sin . . 16 2 1 00 0 433 25 71° = = °− FIG. P35.37
16. 16. 310 Chapter 35 P35.41 Scattered light leaves the center of the photograph (a) in all horizontal directions between θ1 0= ° and 90° from the normal. When it immediately enters the water (b), it is gathered into a fan between 0° and θ2 max given by n n1 1 2 2 2 2 1 00 90 1 333 sin sin . sin . sin θ θ θ θ = = max maax = °48 6. The light leaves the cylinder without deviation, so the viewer only receives light from the center of the photograph when he has turned by an angle less than 48.6°. When the paperweight is turned farther, light at the back surface undergoes total internal reﬂection (c). The viewer sees things outside the globe on the far side. P35.42 Let n x( ) be the index of refraction at distance x below the top of the atmosphere and n x h n( )= = be its value at the planet surface. Then, n x n h x( ) . . = + −⎛ ⎝ ⎞ ⎠ 1 000 1 000 (a) The total time interval required to traverse the atmosphere is ∆t dx n x c dx h h = = ( ) ∫ ∫v0 0 : ∆t c n h x dx h = + −⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥∫ 1 1 000 1 000 0 . . ∆t h c n ch h h c n = + −( )⎛ ⎝⎜ ⎞ ⎠⎟ = +⎛ ⎝⎜ ⎞ ⎠⎟ 1 000 2 1 000 2 2 . . (b) The travel time in the absence of an atmosphere would be h c . Thus, the time in the presence of an atmosphere is n +⎛ ⎝⎜ ⎞ ⎠⎟ 1 000 2 . times larger . P35.43 Let the air and glass be medium 1 and 2, respectively. By Snell’s law, n n2 2 1 1sin sinθ θ= or 1 56 2 1. sin sinθ θ= But the conditions of the problem are such that θ θ1 22= . 1 56 22 2. sin sinθ θ= We now use the double-angle trig identity suggested. 1 56 22 2 2. sin sin cosθ θ θ= or cos . .θ2 1 56 2 0 780= = Thus, θ2 38 7= °. and θ θ1 22 77 5= = °. . q1 max (a) q2 max (b) FIG. P35.41 (c)
17. 17. The Nature of Light and the Laws of Geometric Optics 311 P35.44 (a) ′ = = °θ θ1 1 30 0. n n1 1 2 2sin sinθ θ= 1 00 30 0 1 55 18 8 2 2 . sin . . sin . ° = = ° θ θ (b) ′ = = °θ θ1 1 30 0. θ θ 2 1 1 1 2 1 1 55 30 0 1 = ⎛ ⎝⎜ ⎞ ⎠⎟ = °⎛ − − sin sin sin . sin . n n ⎝⎝⎜ ⎞ ⎠⎟ = °50 8. (c), (d) The other entries are computed similarly, and are shown in the table below. (c) air into glass, angles in degrees (d) glass into air, angles in degrees incidence reﬂection refraction incidence reﬂection refraction 0 0 0 0 0 0 10.0 10.0 6.43 10.0 10.0 15.6 20.0 20.0 12.7 20.0 20.0 32.0 30.0 30.0 18.8 30.0 30.0 50.8 40.0 40.0 24.5 40.0 40.0 85.1 50.0 50.0 29.6 50.0 50.0 none* 60.0 60.0 34.0 60.0 60.0 none* 70.0 70.0 37.3 70.0 70.0 none* 80.0 80.0 39.4 80.0 80.0 none* 90.0 90.0 40.2 90.0 90.0 none* *total internal reﬂection P35.45 For water, sinθc = = 1 4 3 3 4 Thus θc = ( ) = °− sin . .1 0 750 48 6 and d c= ( )⎡⎣ ⎤⎦2 1 00. tanm θ d = ( ) ° =2 00 48 6 2 27. tan . .m m P35.46 (a) We see the Sun moving from east to west across the sky. Its angular speed is ω θ π = = = × −∆ ∆t 2 7 27 10 5rad 86 400 s rad s. The direction of sunlight crossing the cell from the window changes at this rate, moving on the opposite wall at speed v = = × = × =− − rω ( . )( . ) .2 37 7 27 10 1 72 105 4 m rad s m s 00 172. mm s (b) The mirror folds into the cell the motion that would occur in a room twice as wide: v = = ( ) =rω 2 0 174 0 345. .mm s mm s (c), (d) As the Sun moves southward and upward at 50.0°, we may regard the corner of the window as ﬁxed, and both patches of light move northward and downward at 50.0° . FIG. P35.44 FIG. P35.45
18. 18. 312 Chapter 35 P35.47 Horizontal light rays from the setting Sun pass above the hiker. The light rays are twice refracted and once reﬂected, as in Figure (b). The most intense light reaching the hiker, that which represents the visible rainbow, is located between angles of 40° and 42° from the hiker’s shadow. The hiker sees a greater percentage of the violet inner edge, so we consider the red outer edge. The radius R of the circle of droplets is R = ( ) ° =8 00 42 0 5 35. sin . .km km Then the angle f, between the vertical and the radius where the bow touches the ground, is given by cos . . .φ = = = 2 00 2 00 0 374 km km 5.35 kmR or φ = °68 1. The angle ﬁlled by the visible bow is 360 2 68 1 224° − × °( ) = °. so the visible bow is 224 360 62 2 ° ° = . % of a circle . P35.48 (a) 45 0. ° as shown in the ﬁrst ﬁgure to the right. (b) Yes If grazing angle is halved, the number of reﬂections from the side faces is doubled. P35.49 As the beam enters the slab, 1 00 50 0 1 48 2. sin . . sin° = θ giving θ2 31 2= °. The beam then strikes the top of the slab at x1 1 55 31 2 = ° . tan . mm from the left end. Thereafter, the beam strikes a face each time it has traveled a distance of 2 1x along the length of the slab. Since the slab is 420 mm long, the beam has an additional 420 1mm − x to travel after the ﬁrst reﬂection. The number of additional reﬂections is 420 2 1 55 31 2 3 10 1 1 mm 420 mm mm mm − = − °x x . tan . . taan . . 31 2 81 5 ° = or 81 reﬂections since the answer must be an integer. The total number of reﬂections made in the slab is then 82 . Figure (a) Figure (b) FIG. P35.47 FIG. P35.48 FIG. P35.49
19. 19. The Nature of Light and the Laws of Geometric Optics 313 P35.50 Light passing the top of the pole makes an angle of incidence φ θ1 90 0= ° −. . It falls on the water surface at distance from the pole s L d 1 = − tanθ and has an angle of refraction φ2 from 1 00 1 2. sin sinφ φ= n Then s d2 2= tanφ and the whole shadow length is s s L d d n 1 2 1 1 + = − + ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ − tan tan sin sin θ φ s s L d d n 1 2 1 2 + = − + ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ = − tan tan sin cos . θ θ 000 40 0 2 00 40 0 1 33 1m m tan . . tan sin cos . .° + ( ) °⎛− ⎝⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ = 3 79. m P35.51 Deﬁne n1 to be the index of refraction of the surrounding medium and n2 to be that for the prism material. We can use the critical angle of 42.0° to ﬁnd the ratio n n 2 1 : n n2 142 0 90 0sin . sin .° = ° So, n n 2 1 1 42 0 1 49= ° = sin . . Call the angle of refraction θ2 at the surface 1. The ray inside the prism forms a triangle with surfaces 1 and 2, so the sum of the interior angles of this triangle must be 180°. Thus, 90 0 60 0 90 0 42 0 1802. . . .° −( )+ ° + ° − °( ) = °θ Therefore, θ2 18 0= °. Applying Snell’s law at surface 1, n n1 1 2 18 0sin sin .θ = ° sin sin . sin .θ θ1 2 1 2 1 49 18 0= ⎛ ⎝⎜ ⎞ ⎠⎟ = ° n n θ1 27 5= °. *P35.52 (a) As the mirror turns through angle θ, the angle of incidence increases by θ and so does the angle of reﬂection. The incident ray is station- ary, so the reﬂected ray turns through angle 2θ. The angular speed of the reﬂected ray is 2ωm. The speed of the dot of light on the circular wall is 2 2 35ωm R = ( rad/s)(3 m) = 210 m/s . (b) The two angles marked θ in the ﬁgure to the right are equal because their sides are perpendicular, right side to right side and left side to left side. We have cosθ = + = d x d ds dx2 2 and ds dt x dm= +2 2 2 ω . So dx dt ds dt x d d x d d x m= + = + = +2 2 2 2 2 2 2 35 3 ω ( ( rad/s) mm) m) m m s 2 2 2 3 23 3 9 ( .= + ⋅ x FIG. P35.50 FIG. P35.52 FIG. P35.51 continued on next page
20. 20. 314 Chapter 35 (c) The minimum value is 210 m/s for x = 0, just as for the circular wall. (d) The maximum speed goes to inﬁnity as x goes to infinity , which happens when the mirror turns by 45°. (e) To turn by 45° takes the time interval (p/4 rad)/(35 rad/s) = 22.4 ms . *P35.53 (a) For polystyrene surrounded by air, internal reﬂection requires θ3 1 1 00 1 49 42 2= ⎛ ⎝⎜ ⎞ ⎠⎟ = °− sin . . . Then from geometry, θ θ2 390 0 47 8= ° − = °. . From Snell’s law, sin . sin . .θ1 1 49 47 8 1 10= ° = This has no solution. Therefore, total internal reﬂection always happens or the greatest angle is 90°. (b) For polystyrene surrounded by water, θ3 1 1 33 1 49 63 2= ⎛ ⎝⎜ ⎞ ⎠⎟ = °− sin . . . and θ2 26 8= °. From Snell’s law, θ1 30 3= °. (c) No internal refraction is possible since the beam is initially traveling in a medium of lower index of refraction. No angle exists. *P35.54 (a) The optical day is longer. Incoming sunlight is refracted downward at the top of the atmosphere, so an observer can see the rising Sun when it is still geo- metrically below the horizon. Light from the setting Sun reaches her after the Sun is below the horizon geometrically. (b) The picture illustrates optical sunrise. At the center of the Earth, cos . . φ φ θ = × × + = ° = − 6 37 10 8 614 2 98 90 6 2 m 6.37 10 m6 22 98 87 0. .° = ° At the top of the atmosphere n n1 1 2 2 1 1 1 1 000 293 87 0 87 sin sin sin . sin . θ θ θ θ = = ° = ..4° Deviation upon entry is δ θ θ δ = − = ° − ° = ° 1 2 87 364 87 022 0 342. . . Sunrise of the optical day is before geometric sunrise by 0 342 86 400 82 2. .° ° ⎛ ⎝⎜ ⎞ ⎠⎟ = s 360 s. Optical sunset occurs later too, so the optical day is longer by 164 s . FIG. P35.53 FIG. P35.54 1 2 q1 q2 d f
21. 21. The Nature of Light and the Laws of Geometric Optics 315 P35.55 tan . θ1 4 00 = cm h and tan . θ2 2 00 = cm h tan . tan . tan2 1 2 2 2 22 00 4 00θ θ θ= ( ) = sin sin . sin sin 2 1 2 1 2 2 2 21 4 00 1 θ θ θ θ− = − ⎛ ⎝⎜ ⎞ ⎠⎟ (1) Snell’s law in this case is: n n1 1 2 2sin sinθ θ= sin . sinθ θ1 21 333= Squaring both sides, sin . sin2 1 2 21 777θ θ= (2) Substituting (2) into (1), 1 777 1 1 777 4 00 1 2 2 2 2 2 2 2 2 . sin . sin . sin sin θ θ θ θ− = − ⎛⎛ ⎝⎜ ⎞ ⎠⎟ Deﬁning x = sin2 θ, 0 444 1 1 777 1 1 . .− = −x x Solving for x, 0 444 0 444 1 1 777. . .− = −x x and x = 0 417. From x we can solve for θ2: θ2 1 0 417 40 2= = °− sin . . Thus, the height is h = = ° = 2 00 2 00 40 2 2 36 2 . tan . tan . . cm cm cm θ P35.56 δ θ θ= − = °1 2 10 0. and n n1 1 2 2sin sinθ θ= with n1 1= , n2 4 3 = Thus, θ θ θ1 1 2 2 1 2 1 10 0= ( ) = − °( )⎡⎣ ⎤⎦ − − sin sin sin sin .n n (You can use a calculator to home in on an approximate solution to this equation, testing different values of θ1 until you ﬁnd that θ1 36 5= °. . Alternatively, you can solve forθ1 exactly, as shown below.) We are given that sin sin .θ θ1 1 4 3 10 0= − °( ) This is the sine of a difference, so 3 4 10 0 10 01 1 1sin sin cos . cos sin .θ θ θ= ° − ° Rearranging, sin . cos cos . sin10 0 10 0 3 4 1 1° = ° − ⎛ ⎝⎜ ⎞ ⎠⎟θ θ sin . cos . . tan 10 0 10 0 0 750 1 ° ° − = θ and θ1 1 0 740 36 5= ( ) = °− tan . . FIG. P35.55
22. 22. 316 Chapter 35 P35.57 Observe in the sketch that the angle of incidence at point P is g, and using triangle OPQ: sinγ = L R Also, cos sinγ γ= − = − 1 2 2 2 R L R Applying Snell’s law at point P, 1 00. sin sinγ φ= n . Thus, sin sin φ γ = = n L nR and cos sinφ φ= − = − 1 2 2 2 2 n R L nR From triangle OPS, φ α γ+ + °( )+ ° −( ) = °90 0 90 0 180. . or the angle of incidence at point S is α γ φ= − . Then, applying Snell’s law at point S gives 1 00. sin sin sinθ α γ φ= = −( )n n or sin sin cos cos sinθ γ φ γ φ= −[ ]= ⎛ ⎝⎜ ⎞ ⎠⎟ − n n L R n R L n 2 2 2 RR R L R L nR − − ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 2 sinθ = − − −( )L R n R L R L2 2 2 2 2 2 and θ = − − −( )⎡ ⎣⎢ ⎤ ⎦⎥ − sin 1 2 2 2 2 2 2L R n R L R L *P35.58 (a) In the textbook ﬁgure we have r a x1 2 2 = + and r b d x2 2 2 = + −( ) . The speeds in the two media are v v1 1 2 2ϭ ϭc n c nand so the travel time for the light from P to Q is indeed t r r n a x c n b d x c = + = + + + −( )1 1 2 2 1 2 2 2 2 2 v v (b) Now dt dx n c a x x n c b d x= +( ) + + −( )− −1 2 2 1 2 2 2 2 1 2 2 2 2 2( ) (( )( )d x− − =1 0 is the requirement for minimal travel time, which simpliﬁes to n x a x n d x b d x 1 2 2 2 2 2 + = −( ) + −( ) . (c) Now sinθ θ1 2 2 2 2 = + = − + −( ) x a x d x b d x and sin 2 so we have directly n n1 1 2 2sin sinθ θ= . P35.59 To derive the law of reﬂection, locate point O so that the time of travel from point A to point B will be minimum. The total light path is L a b= +sec secθ θ1 2. The time of travel is t a b= ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ +( ) 1 1 2 v sec secθ θ . If point O is displaced by dx, then dt a d b d= ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ +( )= 1 1 1 1 2 2 2 v sec tan sec tanθ θ θ θ θ θ 00 (1) (since for minimum time dt = 0). FIG. P35.57 continued on next page FIG. P35.59
23. 23. The Nature of Light and the Laws of Geometric Optics 317 Also, c d a b+ = + =tan tanθ θ1 2 constant so, a d b dsec sec2 1 1 2 2 2 0θ θ θ θ+ = (2) Divide equations (1) and (2) to ﬁnd θ θ1 2= . P35.60 As shown in the sketch, the angle of incidence at point A is: θ = ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ =− − sin sin .1 12 1 00 30 d R m 2.00 m ..0° If the emerging ray is to be parallel to the incident ray, the path must be symmetric about the centerline CB of the cylinder. In the isosceles triangle ABC, γ α= and β θ= ° −180 Therefore, α β γ+ + = °180 becomes 2 180 180α θ+ ° − = ° or α θ = = ° 2 15 0. Then, applying Snell’s law at point A, nsin . sinα θ= 1 00 or n = = ° ° = sin sin sin . sin . . θ α 30 0 15 0 1 93 P35.61 (a) The apparent radius of the glowing sphere is R3 as shown. For it sin sin sin sin θ θ θ θ 1 1 2 2 3 2 1 2 1 2 3 2 1 = = = = R R R R n n R R R R RR nR3 1= (b) If nR R1 2> , then sinθ2 cannot be equal to nR R 1 2 . The ray considered in part (a) undergoes total internal reﬂection. In this case a ray escaping the atmosphere as shown here is responsible for the apparent radius of the glowing sphere and R R3 2= . FIG. P35.60 R3 FIG. P35.61 (b) R1 R2 R3 q1 q2 FIG. P35.61 (a)
24. 24. 318 Chapter 35 P35.62 (a) At the boundary of the air and glass, the critical angle is given by sinθc n = 1 Consider the critical ray PBB′: tanθc d t = 4 or sin cos θ θ c c d t = 4 . Squaring the last equation gives: sin cos sin sin 2 2 2 2 2 1 4 θ θ θ θ c c c c d t = − = ⎛ ⎝⎜ ⎞ ⎠⎟ . Since sinθc n = 1 , this becomes 1 1 42 2 n d t− = ⎛ ⎝⎜ ⎞ ⎠⎟ or n t d = + ⎛ ⎝⎜ ⎞ ⎠⎟1 4 2 (b) Solving for d, d t n = − 4 12 . Thus, if n = 1 52. and t = 0 600. cm, d = ( ) ( ) − = 4 0 600 1 52 1 2 10 2 . . . cm cm (c) Since violet light has a larger index of refraction, it will lead to a smaller critical angle and the inner edge of the white halo will be tinged with violet light. P35.63 (a) Given that θ1 45 0= °. and θ2 76 0= °. Snell’s law at the ﬁrst surface gives nsin . sin .α = °1 00 45 0 (1) Observe that the angle of incidence at the second surface is β α= ° −90 0. . Thus, Snell’s law at the second surface yields n nsin sin . . sin .β α= ° −( ) = °90 0 1 00 76 0 or n cos sin .α = °76 0 . (2) Dividing Equation (1) by Equation (2), tan sin . sin . .α = ° ° = 45 0 76 0 0 729 or α = °36 1. Then, from Equation (1), n = ° = ° ° = sin . sin sin . sin . . 45 0 45 0 36 1 1 20 α (b) From the sketch, observe that the distance the light travels in the plastic is d L = sinα . Also, the speed of light in the plastic is v = c n , so the time required to travel through the plastic is ∆t d nL c = = = ( ) ×( )v sin . . . sinα 1 20 0 500 3 00 108 m m s 336 1 3 40 10 3 409 . . . ° = × =− s ns t n P B B' qc air d d/4 FIG. P35.62 FIG. P35.63
25. 25. The Nature of Light and the Laws of Geometric Optics 319 *P35.64 sin sin sin sin θ θ θ θ1 2 1 2 0.174 0.342 0.500 0.643 0.7666 0.866 0.940 0.985 0.131 0.261 0.379 0.480 0.576 00.647 0.711 0.740 1.330 4 1.312 9 1.317 7 1.338 5 1.3228 9 1.339 0 1.322 0 1.331 5 The straightness of the graph line demonstrates Snell’s proportionality of the sine of the angle of refraction to the sine of the angle of incidence. The slope of the line is n = ±1 327 6 0 01. . The equation sin sinθ θ1 2= n shows that this slope is the index of refraction, n = ±1 328 0 8. . % *P35.65 Consider an insulated box with the imagined one-way mirror forming one face, installed so that 90% of the electromagnetic radiation incident from the outside is transmitted to the inside and only a lower percentage of the electromagnetic waves from the inside make it through to the out- side. Suppose the interior and exterior of the box are originally at the same temperature. Objects within and without are radiating and absorbing electromagnetic waves. They would all maintain constant temperature if the box had an open window. With the glass letting more energy in than out, the interior of the box will rise in temperature. But this is impossible, according to Clausius’s statement of the second law. This reduction to a contradiction proves that it is impossible for the one-way mirror to exist. ANSWERS TO EVEN PROBLEMS P35.2 227 Mm s P35.4 (a) and (b) See the solution. (c) 0.055 7° P35.6 See the solution. P35.8 27.1 ns P35.10 15.4°, 2.56 m. The light wave slows down as it moves from air to water but the sound wave speeds up by a large factor. The light wave bends toward the normal and its wavelength shortens, but the sound wave bends away from the normal and its wavelength increases. P35.12 (a) 474 THz (b) 422 nm (c) 200 Mm s P35.14 (a) 2.0 × 108 m/s (b) 474 THz (c) 4.2 × 10−7 m P35.16 6.30 cm P35.18 3 39. m FIG. P35.64
26. 26. 320 Chapter 35 P35.20 106 ps P35.22 23.1° P35.24 ∼10−11 s; between 103 and 104 wavelengths P35.26 (a) 41 5. ° (b) 18 5. ° (c) 27 6. ° (d) 42 6. ° P35.28 sin sin cos− − −( )1 2 1n Φ Φ P35.30 See the solution. P35.32 0.171° P35.34 62.5° P35.36 67.2° P35.38 (a) 10.7° (b) air (c) Sound falling on the wall from most directions is 100% reﬂected. P35.40 25.7° P35.42 (a) h c n +⎛ ⎝⎜ ⎞ ⎠⎟ 1 2 (b) larger by n +1 2 times P35.44 (a) See the solution; the angles are ′ = °θ1 30 0. and θ2 18 8= °. (b) See the solution; the angles are ′ = °θ1 30 0. and θ2 50 8= °. . (c) and (d) See the solution. P35.46 (a) 0.172 mm/s (b) 0.345 mm/s (c) Northward at 50.0° below the horizontal (d) Northward at 50.0° below the horizontal P35.48 (a) 45.0° (b) Yes; see the solution. P35.50 3.79 m P35.52 (a) 210 m/s (b) 23.3 (x2 + 9 m2 )/m⋅s (c) 210 m/s for x = 0; this is the same as the speed on the circular wall. (d) The speed goes to inﬁnity as x goes to inﬁnity, when the mirror turns through 45° from the x = 0 point. (e) 22.4 ms P35.54 (a) The optical day is longer. Incoming sunlight is refracted downward at the top of the atmosphere, so an observer can see the rising Sun when it is still geometrically below the horizon. Light from the setting Sun reaches her after the Sun is already below the horizon geometrically. (b) 164 s P35.56 36.5° P35.58 See the solution. P35.60 1.93 P35.62 (a) n = [1 + (4t/d)2 ]1/2 (b) 2.10 cm (c) violet P35.64 See the solution; n = ±1 328 0 8. . %.